Question

the base BC of equilateral triangle ABC lies on y axis .the coordinates of point C are ( 0 , -3) the origin is the midpoint of the base. find the coordinates of the points A and B also find the coordinates of another point D such that be a BACD is a rhombus

Answer 1

Given base BC of equilateral triangle ABC is lies on y - axis , 
And
Coordinates of C ( 0 , -  3 )  and Origin is the mid point of line BC , So, Coordinates of B ( 0 , 3 )

And Point A and D lies of x axis because we know axis are perpendicular to each other , So at origin x axis is perpendicular on y axis .
And
We know in equilateral triangle altitude also bisect opposite side . So we get Coordinates of A (  x  ,  0 ) 
And we know Diagonals of rhombus are perpendicular bisectors to each other , So
Coordinate of D ( - x  , 0 )

As :

 we know
Distance formula d  = (x2 −  x1)2 + (y2 − y1)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√
So,
Side of BC  = (0 −  0)2 + ( − 3 − 3 )2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√ = 0 + (−6)2‾‾‾‾‾‾‾‾‾‾√ = 36‾‾‾√ =  6 unit
As given ABC is equilateral triangle , So

AB  =  BC  =  CA , So

AB  = (0 − x)2 + ( 3 − 0)2‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾‾√ = (x)2 + ( 3 )2‾‾‾‾‾‾‾‾‾‾‾‾√  = x2 + 9‾‾‾‾‾‾‾√  unit , So
we get
x2 + 9‾‾‾‾‾‾‾√ = 6taking whole square on both hand side , we get ⇒x2 + 9 = 36⇒x2 = 25⇒x = 5  unit
So,
Coordinates of A ( 5 , 0 )  and Coordinates of D ( - 5 , 0  )                                   ( Ans )